009C Sample Midterm 1, Problem 2 - Revision history

Kayla Murray at 05:52, 12 November 2017

2017-11-12T05:52:29Z

Kayla Murray at 01:00, 5 November 2017

2017-11-05T01:00:45Z

Kayla Murray at 01:50, 27 February 2017

2017-02-27T01:50:27Z

Kayla Murray at 00:16, 19 February 2017

2017-02-19T00:16:47Z

Kayla Murray at 16:36, 14 February 2017

2017-02-14T16:36:29Z

Kayla Murray at 16:28, 14 February 2017

2017-02-14T16:28:29Z

Kayla Murray at 23:06, 12 February 2017

2017-02-12T23:06:54Z

Kayla Murray at 21:10, 12 February 2017

2017-02-12T21:10:08Z

Kayla Murray at 21:00, 12 February 2017

2017-02-12T21:00:32Z

Kayla Murray at 21:00, 12 February 2017

2017-02-12T21:00:18Z

@@ Line 5: / Line 5: @@
 <span class="exam">(b) Compute &nbsp;<math style="vertical-align: -11px">\lim_{n\rightarrow \infty} s_n.</math>
 <hr>
-(insert picture of handwritten solution)
+[[009C Sample Midterm 1, Problem 2 Solution|'''<u>Solution</u>''']]
 [[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

← Older revision		Revision as of 01:00, 5 November 2017
Line 4:		Line 4:

	<span class="exam">(b) Compute  <math style="vertical-align: -11px">\lim_{n\rightarrow \infty} s_n.</math>		<span class="exam">(b) Compute  <math style="vertical-align: -11px">\lim_{n\rightarrow \infty} s_n.</math>
		+	<hr>
		+	(insert picture of handwritten solution)

		+	[[009C Sample Midterm 1, Problem 2 Detailed Solution\|'''<u>Detailed Solution for this Problem</u>''']]

−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Foundations:  ~~
−	\|-
−	\|The  <math style="vertical-align: 0px">n</math>th partial sum,  <math style="vertical-align: -3px">s_n</math>  for a series  <math>\sum_{n=1}^\infty a_n </math>  is defined as
−	\|-
−	\|
−	~~        <math>s_n=\sum_{i=1}^n a_i.</math>~~
−	\|}
−
−
−	~~'''Solution:'''~~
−
−	~~'''(a)'''~~
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 1:  ~~
−	\|-
−	~~\|We need to find a pattern for the partial sums in order to find a formula.~~
−	\|-
−	~~\|We start by calculating  <math style="vertical-align: -3px">s_2.</math>  We have~~
−	\|-
−	~~\|        <math>s_2=2\bigg(\frac{1}{2^2}-\frac{1}{2^3}\bigg).</math>~~
−	\|}
−
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 2:  ~~
−	\|-
−	~~\|Next, we calculate  <math style="vertical-align: -3px">s_3</math>  and  <math style="vertical-align: -3px">s_4.</math>  We have~~
−	\|-
−	~~\|        <math>\begin{array}{rcl}~~
−	~~\displaystyle{s_3} & = & \displaystyle{2\bigg(\frac{1}{2^2}-\frac{1}{2^3}\bigg)+2\bigg(\frac{1}{2^3}-\frac{1}{2^4}\bigg)}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{2\bigg(\frac{1}{2^2}-\frac{1}{2^4}\bigg)}~~
−	~~\end{array}</math>~~
−	\|-
−	~~\|and~~
−	\|-
−	~~\|        <math>\begin{array}{rcl}~~
−	~~\displaystyle{s_4} & = & \displaystyle{2\bigg(\frac{1}{2^2}-\frac{1}{2^3}\bigg)+2\bigg(\frac{1}{2^3}-\frac{1}{2^4}\bigg)+2\bigg(\frac{1}{2^4}-\frac{1}{2^5}\bigg)}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{2\bigg(\frac{1}{2^2}-\frac{1}{2^5}\bigg).}~~
−	~~\end{array}</math>~~
−	\|}
−
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 3:  ~~
−	\|-
−	~~\|If we look at  <math style="vertical-align: -4px">s_2,s_3,</math>  and  <math style="vertical-align: -4px">s_4, </math>  we notice a pattern.~~
−	\|-
−	~~\|From this pattern, we get the formula~~
−	\|-
−	~~\|        <math>s_n=2\bigg(\frac{1}{2^2}-\frac{1}{2^{n+1}}\bigg).</math>~~
−	\|}
−
−	~~'''(b)'''~~
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 1:  ~~
−	\|-
−	~~\|From Part (a), we have~~
−	\|-
−	~~\|        <math>s_n=2\bigg(\frac{1}{2^2}-\frac{1}{2^{n+1}}\bigg).</math>~~
−	\|}
−
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 2:  ~~
−	\|-
−	~~\|We now calculate  <math style="vertical-align: -11px">\lim_{n\rightarrow \infty} s_n.</math>~~
−	\|-
−	~~\|We get~~
−	\|-
−	~~\|        <math>\begin{array}{rcl}~~
−	~~\displaystyle{\lim_{n\rightarrow \infty} s_n} & = & \displaystyle{\lim_{n\rightarrow \infty} 2\bigg(\frac{1}{2^2}-\frac{1}{2^{n+1}}\bigg)}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\frac{2}{2^2}}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\frac{1}{2}.}~~
−	~~\end{array}</math>~~
−	\|}
−
−
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Final Answer:  ~~
−	\|-
−	~~\|    '''(a)'''     <math>s_n=2\bigg(\frac{1}{2^2}-\frac{1}{2^{n+1}}\bigg)</math>~~
−	\|-
−	~~\|    '''(b)'''     <math>\frac{1}{2}</math>~~
−	\|}
	[[009C_Sample_Midterm_1\|'''<u>Return to Sample Exam</u>''']]		[[009C_Sample_Midterm_1\|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
-<span class="exam">Consider the infinite series <math>\sum_{n=2}^\infty 2\bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg).</math>
+<span class="exam">Consider the infinite series &nbsp;<math>\sum_{n=2}^\infty 2\bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg).</math>
-<span class="exam">(a) Find an expression for the <math style="vertical-align: 0px">n</math>th partial sum <math style="vertical-align: -3px">s_n</math> of the series.
+<span class="exam">(a) Find an expression for the &nbsp;<math style="vertical-align: 0px">n</math>th partial sum &nbsp;<math style="vertical-align: -3px">s_n</math>&nbsp; of the series.
-<span class="exam">(b) Compute <math style="vertical-align: -11px">\lim_{n\rightarrow \infty} s_n.</math>
+<span class="exam">(b) Compute &nbsp;<math style="vertical-align: -11px">\lim_{n\rightarrow \infty} s_n.</math>
@@ Line 9: / Line 9: @@
 !Foundations: &nbsp;
 |-
-|The <math style="vertical-align: 0px">n</math>th partial sum, <math style="vertical-align: -3px">s_n</math> for a series <math>\sum_{n=1}^\infty a_n </math> is defined as
+|The &nbsp;<math style="vertical-align: 0px">n</math>th partial sum, &nbsp;<math style="vertical-align: -3px">s_n</math>&nbsp; for a series &nbsp;<math>\sum_{n=1}^\infty a_n </math>&nbsp; is defined as
 |-
+|
@@ Line 24: / Line 24: @@
 |We need to find a pattern for the partial sums in order to find a formula.
 |-
-|We start by calculating <math style="vertical-align: -3px">s_2</math>. We have
+|We start by calculating &nbsp;<math style="vertical-align: -3px">s_2.</math>&nbsp; We have
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>s_2=2\bigg(\frac{1}{2^2}-\frac{1}{2^3}\bigg).</math>
@@ Line 32: / Line 32: @@
 !Step 2: &nbsp;
 |-
-|Next, we calculate <math style="vertical-align: -3px">s_3</math> and <math style="vertical-align: -3px">s_4.</math> We have
+|Next, we calculate &nbsp;<math style="vertical-align: -3px">s_3</math>&nbsp; and &nbsp;<math style="vertical-align: -3px">s_4.</math>&nbsp; We have
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
@@ Line 52: / Line 52: @@
 !Step 3: &nbsp;
 |-
-|If we look at <math style="vertical-align: -4px">s_2,s_3,</math> and <math style="vertical-align: -4px">s_4, </math> we notice a pattern.
+|If we look at &nbsp;<math style="vertical-align: -4px">s_2,s_3,</math>&nbsp; and &nbsp;<math style="vertical-align: -4px">s_4, </math>&nbsp; we notice a pattern.
 |-
 |From this pattern, we get the formula
@@ Line 71: / Line 71: @@
 !Step 2: &nbsp;
 |-
-|We now calculate <math style="vertical-align: -11px">\lim_{n\rightarrow \infty} s_n.</math>
+|We now calculate &nbsp;<math style="vertical-align: -11px">\lim_{n\rightarrow \infty} s_n.</math>
 |-
 |We get

@@ Line 1: / Line 1: @@
 <span class="exam">Consider the infinite series <math>\sum_{n=2}^\infty 2\bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg).</math>
-::<span class="exam">a) Find an expression for the <math style="vertical-align: 0px">n</math>th partial sum <math style="vertical-align: -3px">s_n</math> of the series.
+<span class="exam">(a) Find an expression for the <math style="vertical-align: 0px">n</math>th partial sum <math style="vertical-align: -3px">s_n</math> of the series.
-::<span class="exam">b) Compute <math style="vertical-align: -11px">\lim_{n\rightarrow \infty} s_n.</math>

@@ Line 1: / Line 1: @@
 <span class="exam">Consider the infinite series <math>\sum_{n=2}^\infty 2\bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg).</math>
-::<span class="exam">a) Find an expression for the <math>n</math>th partial sum <math>s_n</math> of the series.
+::<span class="exam">a) Find an expression for the <math style="vertical-align: 0px">n</math>th partial sum <math style="vertical-align: -3px">s_n</math> of the series.
-::<span class="exam">b) Compute <math>\lim_{n\rightarrow \infty} s_n.</math>
+::<span class="exam">b) Compute <math style="vertical-align: -11px">\lim_{n\rightarrow \infty} s_n.</math>
@@ Line 8: / Line 8: @@
 !Foundations: &nbsp;
 |-
-|The <math style="vertical-align: 0px">n</math>th partial sum, <math style="vertical-align: -3px">s_n</math> for a series <math>\sum_{n=1}^\infty a_n </math>
+|The <math style="vertical-align: 0px">n</math>th partial sum, <math style="vertical-align: -3px">s_n</math> for a series <math>\sum_{n=1}^\infty a_n </math> is defined as
 |-
+|
@@ Line 25: / Line 23: @@
 |We need to find a pattern for the partial sums in order to find a formula.
 |-
-|We start by calculating <math>s_2</math>. We have
+|We start by calculating <math style="vertical-align: -3px">s_2</math>. We have
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>s_2=2\bigg(\frac{1}{2^2}-\frac{1}{2^3}\bigg).</math>
@@ Line 33: / Line 31: @@
 !Step 2: &nbsp;
 |-
-|Next, we calculate <math>s_3</math> and <math>s_4.</math> We have
+|Next, we calculate <math style="vertical-align: -3px">s_3</math> and <math style="vertical-align: -3px">s_4.</math> We have
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
@@ Line 53: / Line 51: @@
 !Step 3: &nbsp;
 |-
-|If we look at <math>s_2,s_3,s_4, </math> we notice a pattern.
+|If we look at <math style="vertical-align: -4px">s_2,s_3,</math> and <math style="vertical-align: -4px">s_4, </math> we notice a pattern.
 |-
 |From this pattern, we get the formula
@@ Line 72: / Line 70: @@
 !Step 2: &nbsp;
 |-
-|We now calculate <math>\lim_{n\rightarrow \infty} s_n.</math>
+|We now calculate <math style="vertical-align: -11px">\lim_{n\rightarrow \infty} s_n.</math>
 |-
 |We get

@@ Line 15: / Line 15: @@
 &nbsp; &nbsp; &nbsp; &nbsp; <math>s_n=\sum_{i=1}^n a_i.</math>
 |}
 '''Solution:'''
@@ Line 83: / Line 84: @@
 \end{array}</math>
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 63: / Line 63: @@
 !Step 1: &nbsp;
 |-
+|From Part (a), we have
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>s_n=2\bigg(\frac{1}{2^2}-\frac{1}{2^{n+1}}\bigg).</math>
 |}
@@ Line 75: / Line 71: @@
 !Step 2: &nbsp;
 |-
+|We now calculate <math>\lim_{n\rightarrow \infty} s_n.</math>
 |-
+|We get
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 |}
@@ Line 89: / Line 89: @@
 |&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>s_n=2\bigg(\frac{1}{2^2}-\frac{1}{2^{n+1}}\bigg)</math>
 |-
-|'''(b)'''
+|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>\frac{1}{2}</math>
 |}
 [[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 87: / Line 87: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)'''
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>s_n=2\bigg(\frac{1}{2^2}-\frac{1}{2^{n+1}}\bigg).</math>
 |-
 |'''(b)'''
 |}
 [[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]